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Given $f: A \subset \mathbb{C} \rightarrow \mathbb{C}$ a holomorphic function, I can represent the function as $f=u+iv, \ u,v:A \rightarrow \mathbb{R}$ and so $f$ can be seen as a function from $A' \subset \mathbb{R}^2$ to $\mathbb{R}^2$ under the identification of $\mathbb{C} \ni z=x+iy$ with $(x,y) \in \mathbb{R}^2$ and of $f$ with $F=(u,v)$.

Now, the problem arise (for me) in two different direction, the first:

A theorem in [Markushevich] ["Theory of functions of a complex variable"]1 says that if we look at $f$ as $F$ (notation above) then $f$ is holomorphic if and only if $F$ is real differentiable and solve Cauchy-Riemann conditions.

Now, this gives us a good method to say when $f$ is holomorphic (at least when we can manage somehow the differentiability of $F$), but (and so the book continues) there is a nice sufficent condition on $F$ which is the continuity of partial derivatives of $u(x,y)$ and $v(x,y)$ (plus C-R equations) to imply holomorphicity of $f$. And that's ok, but this is only a sufficent condition. Wikipedia's page on holomorphic functions seems to do confusion: it says (in "Properties")

If one identifies $\mathbb{C}$ with $\mathbb{R}^2$, then the holomorphic functions coincide with those functions of two real variables with continuous first derivatives which solve the Cauchy–Riemann equations, a set of two partial differential equations.

Is it me, or this is wrong?

Second question: how the fact that $f$ is holomorphic, and so analytical, relates with $u(x,y)$ and $v(x,y)$? The problem, to me, arises when I try to show that a holomorphic function $f=u+iv$ solve the equation $\frac{\partial^2u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$.

i) How can I say that the second (partial) derivatives of $u$ exists? ii) What about the mixed (partial) derivatives of $u$ which are supposed to cancel? How the symmetry requirements on $u$ are satisfied?

In summary: which are some good characterisations of holomorphicity in terms of $u,v$?

Thank you in advance

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  • $\begingroup$ why do you think this is wrong? What (complex) differentiability properties does a holomorphich function have? $\endgroup$ – Roland Feb 17 '17 at 18:52
  • $\begingroup$ because of the "continuity of first derivatives" (which i understood as "continuity of partial derivatives" or continuous differentiation) $\endgroup$ – HaroldF Feb 17 '17 at 18:56
  • $\begingroup$ if $f$ is complex differentiable. what can you say about differentiability of $t \mapsto (f+\overline{f})(t)$ and $t \mapsto (f-\overline{f})(t)$, where $t$ either runs along the real or imaginary line? $\endgroup$ – Roland Feb 17 '17 at 19:02
  • $\begingroup$ I can study the differentiability of $f+\overline{f}$: $u+iv + u - iv= 2u$ and $\overline{f}$ is differentiable (because $-v$ is real differentiable) so $f+\overline{f}$ is differentiable (sum of diff. functions). The same with $f-\overline{f}= 2iv$. So I can say that $u,v$ are real differentiable. I know from the holomorphicity of $f \pm \overline{f}$ that $u$ and $v$ has to be analytic (in a complex sense) $\endgroup$ – HaroldF Feb 17 '17 at 19:23
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I'll take $A\subseteq\mathbb{C}$ to be an open set. A function $f:A\rightarrow \mathbb{C}$ is holomorphic in $A$ if for every $z_{0}\in A$ there exists the limit $$ \lim_{z\rightarrow z_{0}}\frac{f(z)-f(z_{0})}{z-z_{0}}=\ell\in\mathbb{C}, $$ which just says that the derivative $f^{\prime}(z_{0})$ exists in $\mathbb{C}% $. Some books unfortunately require $f^{\prime}$ to be continuous, which just muddles the waters. It is not needed, you get it for free.

Once you know that $f$ is holomorphic, you can prove that $f^{\prime}$ is continuous and it is itself holomorphic, and then you can prove that there exist derivatives of any order and are all holomorphic. To prove this you fix an open ball $B$ in $A$ and prove that for every rectifiable close curve $\gamma$ contained in $B$ you have that $\int_{\gamma}f(z)\,dz=0$. Note that to make sense of the integral you only need $f$ to be continuous and not even differentiability. Cauchy proved this theorem assuming that $f^{\prime}$ was continuous (and this is why some books add this to the definition) but later Goursat proved that the result continues to hold assuming that $f$ is only differentiable.

Once you have $\int_{\gamma}f(z)\,dz=0$ for every rectifiable close curve $\gamma$ contained in $B$ you prove Cauchy's formula, that is, that for every $z_{0}\in B$ \begin{equation} f(z_{0})=\frac{1}{2\pi i}\int_{\partial B(z_{0},r)}\frac{f(z)}{z-z_{0}% }dz\label{cauchy formula}% \end{equation} where the closed ball $\overline{B(z_{0},r)}$ is contained in $B$. The point now is that the right-hand side is a function $h(z_{0})$ which is given by an integral depending on a parameter and as a function of $z_{0}$ you have that $\frac{1}{z-z_{0}}$ is as regular as you want. So you can use theorems on differentiation under the integral sign to conclude that that the right-hand side has derivatives of any order and they are all continuous. In turn the same is true for the left-hand side. So now you know that $f$ has derivatives of any order and they are all continuous. At this point, if you define $u(x,y):=$ real part of $f(x+iy)$ and $v(x,y):=$immaginary part of $f(x+iy)$, then $u$ and $v$ are $C^{\infty}$ because they are given by the composition of the function $(x,y)\mapsto x+iy$ which is $C^{\infty}$ with the function $f$ which is also $C^{\infty}$. In particular you can use the chain rule and take second derivatives of $u$ and $v$ and prove that the Cauchy-Riemann equations are satisfied and that $u$ and $v$ is harmonic (the mixed derivatives cancel out because $u$ and $v$ are $C^{\infty}$). Hope this answers your second questions.

Now the converse implication is not perfect. There is a nice review article of Gray and Morris . One good result is the following. Take $U\subseteq \mathbb{R}^{2}$ be an open set and let $u:U\rightarrow\mathbb{R}$ and $v:U\rightarrow\mathbb{R}$ be such that there exist the partial derivatives $\partial_{x}u$ and $\partial_{y}u$ and $\partial_{x}v$ and $\partial_{y}v$. Note that the existence of partial derivatives DOES not imply that $u$ and $v$ are differentiable in $U$. You also assume that $u$ and $v$ satisfy the Cauchy-Riemann equations.

These hypotheses are not enough, since they do not even imply that $u$ and $v$ are continuous. The function $$ f(z):=\left\{ \begin{array} [c]{ll}% \exp(-z^{-4}) & \text{if }z\neq0\\ 0 & \text{if }z=0, \end{array} \right. $$ does not have a derivative at $z=0$ but the corresponding functions $u$ and $v$ satisfy the Cauchy-Riemann equations in $\mathbb{R}^{2}$.

So you need an extra hypothesis. There are several variants. One that I like is that $u:U\rightarrow\mathbb{R}$ and $v:U\rightarrow\mathbb{R}$ admit $\partial_{x}u$ and $\partial_{y}u$ and $\partial_{x}v$ and $\partial_{y}v$ in $U$, $u$ and $v$ satisfy the Cauchy-Riemann equations, and that the function $f(z):=u(x,y)+iv(x,y)$, where $z=x+iy$, is continuous in the domain $A=\{z=x+iy:\,(x,y)\in U\}$. The idea of the proof is to to mollify the functions $u$ and $v$, taking $u_{\varepsilon}=\varphi_{\varepsilon}\ast u$ and $v_{\varepsilon}=\varphi_{\varepsilon}\ast v$, where $\varphi _{\varepsilon}$ is a nice kernel, and prove that the $C^{\infty}$ functions $u_{\varepsilon}$ and $v_{\varepsilon}$ satisfy the Cauchy-Riemann equations. Using that, you can prove that $\int_{\gamma}f_{\varepsilon}(z)\,dz=0$ for every rectifiable close curve $\gamma$ contained in $B$, where $f_{\varepsilon }(z):=u_{\varepsilon}(x,y)+iv_{\varepsilon}(x,y)$, where $z=x+iy$, and then continue as before to prove that $f_{\varepsilon}$ is analytic. In turn you get the Cauchy formula (1) for $f_{\varepsilon}$ and then since $f$ is continuous, you can let $\varepsilon\rightarrow0$ to prove that $f$ satisfies the Cauchy formula and so is holomorphic. Hope this answers your first question. Read the paper, it is well-written.

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Let $\Omega\subseteq\mathbb{C}$ open, $z_{0}\in\Omega,\ f:\Omega\rightarrow\mathbb{C}$.

  • Theorem: $f$ is complex-differentiable at $z_{0}$ iff it is real-differentiable at $z_{0}$ and satisfies the Cauchy-Riemann equations.

Since being holomorphic in $\Omega$ implies $f\in C^{\infty}(\Omega)$ (but being holomorphic at a single point does not implies to be $C^{\infty}$ at the point), we also have:

  • Corolary: $f$ is complex-differentiable in $\Omega$ iff in $\Omega$ it is real-differentiable, satisfies the Cauchy-Riemann equations and the patial derivatives are continous.

EDIT: Proof of the theorem: $f$ is complex-differentiable at $z_{0}$ iff $\exists a,b\in\mathbb{R}$: $$ \mathbb{lim}_{z\rightarrow z_{0}} \frac{|f(z)-f(z_{0})-(a+ib)(z-z_{0})|}{|z-z_{0}|}=0$$ iff $\exists a,b\in\mathbb{R}$: $$ \mathbb{lim}_{(x,y)\rightarrow (x_{0},y_{0})}\frac{\left | f(x,y)-f(x_{0},y_{0})-\begin{pmatrix} a & -b\\ b & a \end{pmatrix}\left ( \begin{pmatrix} x\\ y \end{pmatrix}-\begin{pmatrix} x_{0}\\ y_{0} \end{pmatrix}\right )\right |}{\left | \begin{pmatrix} x\\ y \end{pmatrix}-\begin{pmatrix} x_{0}\\ y_{0} \end{pmatrix} \right |} =0$$ iff $f$ is real-differentiable and satisfies the Cauchy-Riemann equations.

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    $\begingroup$ Can you give an example for the claim: being holomorphic at a single point does not implies to be $C^{\infty}$ at the point. $\endgroup$ – user315531 Jan 22 at 22:53

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